Gauss's Lemma Proof

Question

Theorem 2.39 (Gauss’s Lemma). A polynomial $f ∈ \mathbb{Z}[x] ⊆ \mathbb{Q}[x]$ of the form $$f(x) = x^n + a_{n−1}x+^{n−1}+ ...+ a_1x + a_0$$ is irreducible in $\mathbb{Q}[x]$ if and only if it is irreducible in $\mathbb{Z}[x]$. More precisely, if $f(x) ∈ \mathbb{Z}[x]$, then $f(x)$ factors into a product of two polynomials of lower degrees $r$ and $s$ in $\mathbb{Q}[x]$ if and only if it has such a factorisation with polynomials of the same degrees $r$ and $s$ in $\mathbb{Z}[x]$.

This is a theorem that we have gotten in our lecture notes, when I search the proof for Gauss's Theorem online I get different theorems to the one above and many different proofs that don't seem right for this theorem. Are there many different Gauss's Lemma?

What would a proof for this version look like? Could someone send me a link to a website that has th eproof for this Lemma?

Answer 1

"Abstract Algebra" of Dummit and Foote (pag. 303) has a detailed proof.

Let $p(x) \in \mathbb{Z}[x]$ such that $p(x)=A(x)B(x)$ for some non-zero degree polynomials $A(x)$ and $B(x)$ in $\mathbb{Q}[x]$. Then we can find $r$ and $s$ in $\mathbb{Q}$ such that $a(x)=rA(x)$ and $b(x)=sB(x)$ are both in $\mathbb{Z}[x]$ and $p(x)=a(x)b(x)$ (i.e. a factorization of $p(x)$ in $\mathbb{Z}[x]$).

Proof.

Let $k$ be the common denominator of all the coefficients of $A(x)$ hence $a'(x)=kA(x)$ is in $\mathbb{Z}[x]$. In the same way we find a $b'(x)=qB(x)$ in $\mathbb{Z}[x]$.Let $d=kq$, we get $dp(x)=a'(x)b'(x)$. If $d=1$ we've done, so suppose $d \neq 1$. Write the factorization of $d$ into, non necessarily distinct, prime numbers $d=p_1p_2...p_n$.

For every $i$ the ring $\mathbb{Z}/ p_i\mathbb{Z}[x]$ is an integer domain where $\overline{a'(x)} \overline{b'(x)}=\overline{0}$, hence one of the two factor must be $\overline{0}$, for example $\overline{a'(x)}$. This implies that all the coefficients of $a'(x)$ are divisible for $p_i$ so that also $\frac{1}p_i a'(x)$ is in $\mathbb{Z}[x]$. In other words we can cancel $p_i$ from both the sides of $dp(x)=a'(x)b'(x)$ and still get an equation of $\mathbb{Z}[x]$. Iterating over all the $p_i$ complete the proof. $r$ is given from $k$ divided all the primes removed from $a'(x)$, in the same way $s$ is given from $q$ divided all the primes removed from $b'(x)$.

Answer 2

There are several related results that are called “Gauss’s Lemma” in the literature. One of the most common versions is:

Theorem. Let $f(x)$ and $g(x)$ be polynomials with integer coefficients. If both $f(x)$ and $g(x)$ are primitive, then $fg$ is primitive.

Other versions assert that the content of a product is the product of the content.

All versions of Gauss’s Lemma lead to the result you are quoting: that primitive polynomials with integer coefficients are irreducible over $\mathbb{Z}$ if and only if they are irreducible over $\mathbb{Q}$, and from there to the proof that if $R$ is a UFD, then $R[x]$ is a UFD.

So it is no surprise that you find different results on-line called “Gauss’s Lemma”. For example, my own paper with “Gauss’s Lemma” in the title uses “Gauss’s Lemma” to refer to the result on the product of primitive polynomials.

Gauss’s own statement of the Lemma is in his Disquisitiones Arithmeticae, and to be honest, is closer to your statement than mine. In fact, it is the contrapositive of the “if” clause of your theorem. It is Article 42 in Section II, page 25 of the Arthur A. Clarke translation published by Yale University Press in 1965. It reads:

If the coefficients $A$, $B$, $C,\ldots,N$; $a,b,c,\ldots,n$ of two functions of the form $$\begin{align*} x^m + Ax^{m-1}+Bx^{m-2}+Cx^{m-3}+\cdots + N \tag{P}\\ x^{\mu} + ax^{\mu-1} + bx^{\mu-2} + cx^{\mu-3} + \cdots + n \tag{Q} \end{align*}$$ are all rational and not all integers, and if the product of (P) and (Q) $$ = x^{m+\mu} + \mathfrak{A}x^{m+\mu-1} + \mathfrak{B}x^{m+\mu-2} + \text{etc.} + \mathfrak{Z}$$ then not all the coefficients $\mathfrak{A}$, $\mathfrak{B},\ldots,\mathfrak{Z}$ can be integers.

This version gives yours; if the polynomial is irreducible over $\mathbb{Q}$, then it is easy to verify it is irreducible over $\mathbb{Z}$ (the only problem would be if the content were not trivial, which is not the case). And conversely, if it is reducible over $\mathbb{Q}$, then by this version of Gauss’s Lemma the factors must have integer coefficients, so it is reducible over $\mathbb{Z}$ as well.

Gauss proves the result. Here is a paraphrase of his proof, generally following his notation (with some modifications):

Gauss’s Proof (paraphrased). Express every coefficient as a fraction in lowest term, and take a prime $p$ that divides at least one of the denominators (possible since not all coefficients are integers. Say it divides a denominator in a coefficient in (P); divide (Q) by $p$. Then at least one coefficient of $\frac{1}{p}$(Q) has $p$ as a factor in the denominator (the leading coefficient, for one). Find the term of largest degree in (P) that contains the largest possible power of $p$ that divides the denominator of a coefficient of (P); say that term is $Gx^g$, and say the denominator is divisible by $p^t$ but no larger power. Similarly, let $\Gamma x^{\gamma}$ be the term of largest degree in $\frac{1}{p}$(Q) whose denominator is divisible by the largest power of $p$ that occurs in any coefficient of $\frac{1}{p}$(Q), with the corresponding power of $p$ being $\tau$. So $t+\tau\geq 2$. The claim is that the term $x^{g+\gamma}$ in the product has a fractional coefficient with denominator divisible by $t+\tau-1$.

Let us denote the terms of (P) that go before $Gx^g$ by ${}_1Gx^{g+1}$, ${}_2Gx^{g+1}$, and so on, and the ones the follow by $G_1x^{g-1}$, $G_2x^{g-2}$, etc. Similarly, the terms that go before $\Gamma x^{\gamma}$ in $\frac{1}{p}$(Q) will be ${}_1\Gamma x^{\gamma+1}$, ${}_2\Gamma x^{\gamma+2}$, and so on, while the ones that go after are $\Gamma_1 x^{\gamma-1}$, $\Gamma_2 x^{\gamma-2}$, etc. Then the coefficient of $x^{g+\gamma}$ in the product of (P) and $\frac{1}{p}$(Q) is $$G\Gamma + \sum_i {}_iG\Gamma_i + \sum_j {}_j\Gamma G_j$$ If we express $G\Gamma$ in lowerst terms, it has denominator divisible by $p^{t+\tau}$. If any of the other terms is a fraction, it has denominators that are divisible by lower powers of $p$, because they each involve one factor with strictly smaller powers of $p$ than $t$ or $\tau$, and one with no larger power of $p$, by the choice of $g$ and $\gamma$. So we can write $$G\Gamma = \frac{e}{fp^{t+\tau}}$$ and the sum of all the others will be of the form $$\frac{e’}{f’p^{t+\tau-\delta}}$$ where $e$, $f$, $e’$, and $f’$ are relatively prime to $p$, and $\delta\gt 0$. The coefficient is then the sum of these, which is $$\frac{ef’ + e’fp^{\delta}}{ff’p^{t+\tau}}.$$ The numerator is not divisible by $p$, so there can be no reduction on the powers of $p$ in the denominator.

That means that the coefficient of $x^{g+\gamma}$ in the product of (P) and (Q) (rather than the product of (P) and $\frac{1}{p}$(Q) that we just computed) is $$\frac{ef’ + e’f p^{\delta}}{ff’p^{t+\tau-1}}$$ whose denominator is divisible by $p$ and is already in lowest terms; that is, a rational, not an integer. $\Box$

Answer 3

Lemma 1:
Let $G(x),H(x)\in\mathbb{Z}[x]$.
Let $F(x) := G(x)H(x)$.
Let $p$ be a prime number.
If $p$ divides all the coefficients of $F(x)$, then $p$ divides all the coefficients of $G(x)$ or $p$ divides all the coefficients of $H(x)$.

Proof:
Let $F(x)=x^n+A_{n−1}x^{n−1}+\dots+A_1x+A_0$.
Let $G(x)=x^r+B_{r−1}x^{r−1}+\dots+B_1x+B_0$.
Let $H(x)=x^s+C_{s−1}x^{s−1}+\dots+C_1x+C_0$.
Let $B_i:=0$ for $i\geq r$.
Let $C_i:=0$ for $i\geq s$.
Assume that $p$ does not divide all the coefficients of $G(x)$ and $p$ does not divide all the coefficients of $H(x)$.
Let $i:=\min\{k\in\{0,1,\dots,r-1\}\mid p\text{ does not devide }B_k\}.$
Let $j:=\min\{k\in\{0,1,\dots,s-1\}\mid p\text{ does not devide }C_k\}.$
Then $A_{i+j}=B_0C_{i+j}+\dots+B_{i-1}C_{j+1}+B_iC_j+B_{i+1}C_{j-1}+\dots+B_{i+j}C_0.$
$B_0C_{i+j}+\dots+B_{i-1}C_{j+1}$ is divisible by $p$ since $p$ divides each of $B_0,\dots,B_{i-1}$.
$B_{i+1}C_{j-1}+\dots+B_{i+j}C_0$ is divisible by $p$ since $p$ divides each of $C_0,\dots,C_{j-1}$.
$A_{i+j}$ is divisible by $p$ by our assumption.
So, $B_iC_j$ is divisible by $p$.
By a famous theorem, $p$ divides $B_i$ or $C_j$.
This is a contradiction.

Gauss' Lemma:
Let $f(x)=x^n+a_{n−1}x^{n−1}+\dots+a_1x+a_0\in\mathbb{Z}[x]$.
Suppose that $f(x)$ is reducible over $\mathbb{Q}$.
Then, there exist non-constant polynomials $G(x),H(x)\in\mathbb{Z}[x]$ such that $f(x)=G(x)H(x)$.

Proof:
Let $g(x),h(x)$ be non-constant elements of $\mathbb{Q}[x]$ such that $f(x)=g(x)h(x)$.
There exists a positive integer $b$ such that $bg(x)\in\mathbb{Z}[x]$.
There exists a positive integer $c$ such that $ch(x)\in\mathbb{Z}[x]$.
Let $a:=bc$.
Let $G(x):=bg(x).$
Let $H(x):=ch(x).$
Then, $af(x)=G(x)H(x)$ holds.
If $a=1$, then Gauss' Lemma holds.
IF $a>1$, then there exists a prime number $p_1$ such that $a=p_1a_1$ for some positive integer $a_1$.
By Lemma 1, $p_1$ divides all the coefficients of $G(x)$ or $p_1$ divides all the coefficients of $H(x)$.
If $p_1$ divides all the coefficients of $G(x)$, let $G(x):=\frac{G(x)}{p_1}$.
If $p_1$ divides all the coefficients of $H(x)$, let $H(x):=\frac{H(x)}{p_1}$.
Then, $a_1f(x)=G(x)H(x)$ holds.
If $a_1=1$, then Gauss' Lemma holds.
IF $a_1>1$, then there exists a prime number $p_2$ such that $a_1=p_2a_2$ for some positive integer $a_2$.
By Lemma 1, $p_2$ divides all the coefficients of $G(x)$ or $p_2$ divides all the coefficients of $H(x)$.
If $p_2$ divides all the coefficients of $G(x)$, let $G(x):=\frac{G(x)}{p_2}$.
If $p_1$ divides all the coefficients of $H(x)$, let $H(x):=\frac{H(x)}{p_2}$.
Then, $a_2f(x)=G(x)H(x)$ holds.
If $a_2=1$, then Gauss' Lemma holds.
IF $a_2>1$, then $\cdots$

Since $a:=a_0>a_1>a_2>\cdots$, there exists a non-negative integer $l$ such that $a_l=1$.
Then, $a_lf(x)=f(x)=G(x)H(x)$ holds.

Answer 4

Since the original question is not about general rings and their quotient fields, there is a much shorter and really beautiful proof of the specific $\mathbb Z[x]$ and $\mathbb Q[x]$ case-

Let the polynomial be $$P(x)=\sum_{k=0}^{n-1}a_kx^k+x^n$$ and let $x=\frac pq$ be a root with $(p,q)=0$ (i.e., $p$ and $q$ are relatively prime). So, $$0=q^nP(p/q)=\sum_{k=0}^{n-1}a_kq^{n-k}p^k+p^n$$ Hence, $q|p^n$ which implies $q=\pm 1$ hence completing the proof.